Vector

Product of the vectors

Owing to the different way in which vectors occur in various physical problems, the product of two vectors \(\overrightarrow {a}\) and \(\overrightarrow {b}\) is defined in the following two ways:

  1. scalar product or dot product \(\overrightarrow a\) . \(\overrightarrow b\) (read as \(\overrightarrow a\) dot \(\overrightarrow b\))
  2. vector product or cross product \(\overrightarrow a\)× \(\overrightarrow b\) (read as \(\overrightarrow a\) cross \(\overrightarrow b\)).

The dot product \(\overrightarrow a\) . \(\overrightarrow b\) gives a scalar result while the cross product \(\overrightarrow a\)× \(\overrightarrow b\) gives a vector result.

Scalar product of two vectors

.

The scalar product of two vectors \(\overrightarrow a\) and \(\overrightarrow b\) is defined as the product of the magnitude of two vectors multiplied by the cosine of the angle \(\theta\) between their directions.

Thus, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos\(\theta\)

where, |\(\overrightarrow a\)| = a and |\(\overrightarrow b\)| = b.

Now,

.

Draw perpendicular BM from B to OA.

Here,

\(\overrightarrow {OA}\) = \(\overrightarrow a\) and \(\overrightarrow {OB}\) = \(\overrightarrow b\)

Now,

\(\begin{align*} \overrightarrow a . \overrightarrow b &= |\overrightarrow a| |\overrightarrow b| cos \theta\\ &= a b cos\theta\\ &= (OA) (OB) cos\theta\\ &= OA (OB cos\theta)\\ &= (OA)(OM)\\ &= (magnitude\;of\;\overrightarrow a) (component\;of\;\overrightarrow b\;in\;the\;direction\;of\;\overrightarrow a)\\ \end{align*}\)

So, it is clear that the scalar product of two vectors is equivalent to the product of the magnitude of one vector with the component of the other vector in the direction of this vector.

If we write \(\overrightarrow a\) . \(\overrightarrow b\), the rotation of \(\overrightarrow a\) towards \(\overrightarrow b\) is anticlockwise and the angle \(\theta\) is taken to be positive.

∴ \(\overrightarrow a\) . \(\overrightarrow b\) = ab cos\(\theta\)

If we write \(\overrightarrow b\) . \(\overrightarrow a\), the rotation of \(\overrightarrow b\) towards \(\overrightarrow a\) is clockwise and the angle \(\theta\) is taken to be negative.

∴ \(\overrightarrow b\) . \(\overrightarrow a\) = b a cos(-\(\theta\)) = ba cos\(\theta\)

Hence, \(\overrightarrow a\) . \(\overrightarrow b\) = \(\overrightarrow b\) . \(\overrightarrow a\)

Thus, scalar product is commulative.

.

Let us consider two points A(a1, a2) and B(b1, b2) in the plane. Then,

position vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\)

position vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\)

Magnitudes of \(\overrightarrow a\) and \(\overrightarrow b\) are

|\(\overrightarrow {OA}\)| = OA = a = |\(\overrightarrow a\)|

|\(\overrightarrow {OB}\)| = OB = b = |\(\overrightarrow b\)|

Let \(\angle\)XOA =β, \(\angle\)XOB =α and \(\angle\)AOB =θ. Then,α -β =θ.

Draw perpendiculars AM and BN from A and B to the x-axis. Then,

OM = a1, MA = a2, ON = b1 and NB = b2.

From the right-angled triangle OMA,

cosβ = \(\frac {OM}{OA}\) = \(\frac {a_1}a\)∴ a1 = a cosβ

sinβ = \(\frac {MA}{OA}\) = \(\frac {a_2}a\)∴ a2 = a sinβ

Similarly,

From the right-angled triangle ONB,

b1 = b cosα and b2 = b sinα

Now,

\(\begin{align*} a_1b_1 + a_2b_2 &= a cosβ b cosα + a sinβ sinα\\ &= ab cos (α - β)\\ &= |\overrightarrow a| |\overrightarrow b| cos \theta...................(i)\\ \end{align*}\)

But,

By the deefinition of scalar product of two vectors \(\overrightarrow a\) and \(\overrightarrow b\),

Now,

From (i) and (ii),

\(\overrightarrow a\) . \(\overrightarrow b\) = a1b1 + a2b2.

This result leads us to define the scalar product of two vectors in another way.

Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) be two vectors. Then the scalar product of \(\overrightarrow a\) and \(\overrightarrow b\) is denoted by \(\overrightarrow a\) . \(\overrightarrow b\) and is defined by \(\overrightarrow a\) . \(\overrightarrow b\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) .\(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) = a1b1 + a2b2.

Again,

From (i),

|\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = a1b1 + a2b2

or, cos\(\theta\) = \(\frac {a_1b_1 + a_2b_2}{|\overrightarrow a| |\overrightarrow b|}\)

This result gives us angle between two vectors \(\overrightarrow a\) and \(\overrightarrow b\).

\(\theta\) = cos-1 \(\frac {a_1b_1 + a_2b_2}{|\overrightarrow a| |\overrightarrow b|}\)

Properties of Scalar Product

The following properties are satisfied by the scalar product of vectors:

Let \(\overrightarrow a\), \(\overrightarrow b\) and \(\overrightarrow c\) be three vectors.

  1. Commulative Property: \(\overrightarrow a\) . \(\overrightarrow b\) = \(\overrightarrow b\) . \(\overrightarrow a\)
  2. Distributive Property: \(\overrightarrow a\) . (\(\overrightarrow b\) + \(\overrightarrow c\)) = \(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow a\) . \(\overrightarrow c\)
  3. Associative Property:m \(\overrightarrow a\) . n \(\overrightarrow b\) = mn (\(\overrightarrow a\) . \(\overrightarrow b\))

Perpendicular Vectors

Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) be two vectors. If \(\overrightarrow a\) and \(\overrightarrow b\) are perpendicular to each other, then the angle between \(\overrightarrow a\) and \(\overrightarrow b\) is \(\theta\) =- 90°.

Now,

\(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 90° = 0

Conversely,

Let \(\overrightarrow a\) . \(\overrightarrow b\) = 0

Then,

|\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = 0

or, ab cos\(\theta\) = 0

or, cos\(\theta\) = 0

∴\(\theta\) = 90°

Thus, if two vectors are perpendicular to each other (or orthogonal), their scalar product is zero.

Parallel Vector:

Let \(\overrightarrow a\) and \(\overrightarrow b\) be two vectors. If \(\overrightarrow a\) and \(\overrightarrow b\) are parallel to each other then the angle between them is 0° or 180°.

Now,

If \(\theta\) = 0°, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 0° = ab

If \(\theta\) = 180°, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 180° = -ab

Thus, two vector \(\overrightarrow a\) and \(\overrightarrow b\) are parallel to each other if \(\overrightarrow a\) . \(\overrightarrow b\) = ab or, \(\overrightarrow a\) . \(\overrightarrow b\) = -ab.

Length of a vector (Modulus of a vector)

Length of a vector (Modulus of a vector)
Length of a vector (Modulus of a vector)

Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) be a plane vector.

Then, \(\overrightarrow a\) . \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) . \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) = a1a1 + a2a2= a12 + a22 = a2

∴ a = \(\sqrt {\overrightarrow a. \overrightarrow a}\)

Hence, the length of a vector \(\overrightarrow a\) is the positive square root of the scalar product \(\overrightarrow a\) . \(\overrightarrow a\).

The scalar product of a vector with itself is often written as the square of the vector.

So,\(\overrightarrow a\) . \(\overrightarrow a\) = a2 i.e. \(\overrightarrow {a^2}\) = a2

Some simple identities

  1. (\(\overrightarrow a\) + \(\overrightarrow b\)) = \(\overrightarrow {a^2}\) + 2\(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow {b}\) = a2 + 2 \(\overrightarrow a\) . \(\overrightarrow b\) + b2
    (\(\overrightarrow a\) + \(\overrightarrow b\)) =(\(\overrightarrow a\) + \(\overrightarrow b\)) .(\(\overrightarrow a\) + \(\overrightarrow b\)) = \(\overrightarrow a\) . \(\overrightarrow a\) + \(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow b\) . \(\overrightarrow a\) + \(\overrightarrow b\) . \(\overrightarrow b\) = \(\overrightarrow a\) + 2\(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow {b^2}\) = a2 + 2\(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow b\)
  2. (\(\overrightarrow a\) - \(\overrightarrow b\))2 = \(\overrightarrow {a^2}\) - 2 \(\overrightarrow a\) . \(\overrightarrow a\) + \(\overrightarrow {a^2}\) = a2 - 2\(\overrightarrow a\) . \(\overrightarrow b\) + b2
    (\(\overrightarrow a\) - \(\overrightarrow b\))2 =(\(\overrightarrow a\) - \(\overrightarrow b\)) .(\(\overrightarrow a\) - \(\overrightarrow b\))= \(\overrightarrow a\) . \(\overrightarrow a\) - \(\overrightarrow a\) . \(\overrightarrow b\) - \(\overrightarrow b\) . \(\overrightarrow a\) + \(\overrightarrow b\) . \(\overrightarrow b\) = \(\overrightarrow a\) - 2\(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow b\) = a2 - 2\(\overrightarrow a\) . \(\overrightarrow b\) + b2
  3. (\(\overrightarrow a\) + \(\overrightarrow b\)) .(\(\overrightarrow a\) - \(\overrightarrow b\) = \(\overrightarrow {a^2}\) - \(\overrightarrow {b^2}\) = a2 - b2
    (\(\overrightarrow a\) + \(\overrightarrow b\)) .(\(\overrightarrow a\) - \(\overrightarrow b\) = \(\overrightarrow a\) . \(\overrightarrow a\) - \(\overrightarrow a\) . \(\overrightarrow b\) + \(\overrightarrow b\) . \(\overrightarrow b\) = \(\overrightarrow {a^2}\) - \(\overrightarrow b\) = a2 - b2

Mutually perpendicular unit vector \(\overrightarrow i\) and \(\overrightarrow j\)

Let OX and OY be two mutually perpendicular straight lines. Then the unit vector along OX and OY denoted by \(\overrightarrow i\) and \(\overrightarrow j\) are defined by \(\overrightarrow i\) = \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) and \(\overrightarrow j\) = \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\).

Now,

\(\overrightarrow i\) . \(\overrightarrow i\) = \(\overrightarrow {i^2}\) =\(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) . \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) = 1 + 0 = 1

\(\overrightarrow j\) . \(\overrightarrow j\) = \(\overrightarrow {j^2}\) =\(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) . \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) = 0 + 1 = 1

\(\overrightarrow i\) . \(\overrightarrow j\) = \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) . \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) = 0 + 0 = 0

\(\overrightarrow j\) . \(\overrightarrow i\) = \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) . \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) = 0 + 0 = 0

The value of scalar product of \(\overrightarrow i\) and \(\overrightarrow j\) can be remembered from the table given alongside.

Representation of a vector in terms of unit vectors

Let \(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) be a vector. It can be written as

\(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) = \(\begin{pmatrix} x\\ 0\\ \end{pmatrix}\) = \(\begin{pmatrix} 0\\ y\\ \end{pmatrix}\) = x\(\begin{pmatrix}1\\ 0\\ \end{pmatrix}\) + y\(\begin{pmatrix}0\\ 1\\ \end{pmatrix}\) = x\(\overrightarrow i\) + y\(\overrightarrow j\).

Similarly,

If \(\overrightarrow a\) = nx + y\(\overrightarrow j\), then:

\(\overrightarrow a\)= x\(\overrightarrow i\) + y\(\overrightarrow j\) = x\(\begin{pmatrix}1\\ 0\\ \end{pmatrix}\) + y\(\begin{pmatrix}0\\ 1\\ \end{pmatrix}\) = \(\begin{pmatrix}x\\ 0\\ \end{pmatrix}\) + \(\begin{pmatrix}0\\ y\\ \end{pmatrix}\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\)

Hence, every plane vector \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) can be represented by x\(\overrightarrow i\) + y\(\overrightarrow j\) and conversely.

Vector operations in terms of unit vectors

Let \(\overrightarrow a\) = a1\(\overrightarrow i\) + a2\(\overrightarrow j\) and \(\overrightarrow b\) = b1\(\overrightarrow i\) + b2\(\overrightarrow j\)

  1. Addition of vectors:
    \(\overrightarrow a\) + \(\overrightarrow b\) = a1\(\overrightarrow i\) + a2\(\overrightarrow j\) + b1\(\overrightarrow i\) + b2\(\overrightarrow j\) = (a1 + b1)\(\overrightarrow i\) + (a2 + b2)\(\overrightarrow j\)
  2. Substraction of Vectors:
    \(\overrightarrow a\) - \(\overrightarrow b\) = a1\(\overrightarrow i\) + a2\(\overrightarrow j\) - (b1\(\overrightarrow i\) + b2\(\overrightarrow j\)) =a1\(\overrightarrow i\) + a2\(\overrightarrow j\) - b1\(\overrightarrow i\) - b2\(\overrightarrow j\)
  3. Scalar product of vectors:
    \(\begin{align*} \overrightarrow a . \overrightarrow b &= (a_1\overrightarrow i + a_2\overrightarrow j) . (b_1\overrightarrow i + b_2\overrightarrow j)\\ &= a_1b_1 \overrightarrow i . \overrightarrow i + a_1b_2 \overrightarrow i . \overrightarrow j + a_2b_1 \overrightarrow j . \overrightarrow i + a_2b_2 \overrightarrow j . \overrightarrow j\\ &= a_1b_1 + 0 + 0 + a+2b_2\\ &= a_1b_1 + a_2b_2\\ \end{align*}

Magnitude and direction of a vector in terms of unit vectors

Let \(\overrightarrow a\) = x\(\overrightarrow i\) + y\(\overrightarrow j\) be a vector.

Then,

\(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\)

∴ X- component of \(\overrightarrow a\) = x and Y-component of \(\overrightarrow a\) = y

Magnitude of \(\overrightarrow a\)

|\(\overrightarrow a\)| = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(coefficient\;of\;\overrightarrow i)^2 + (coefficient\;of\;\overrightarrow j)^2}\)

Direction of \(\overrightarrow a\)

tan\(\theta\) = \(\frac yx\) = \(\frac {coefficient\;of\;\overrightarrow j}{coefficient\;of\;\overrightarrow i}\)

Unit vector along the direction of \(\overrightarrow a\)

\(\widehat a\) = \(\frac {\overrightarrow a}{|\overrightarrow a|}\) = \(\frac {1}{\sqrt {x^2 + y^2}}\) (x\(\overrightarrow i\) + y\(\overrightarrow j\)) = \(\frac {x\overrightarrow i + y\overrightarrow j}{\sqrt {x^2 + y^2}}\)

Vector Geometry

Theorem 1: (Mid- point Formula)

If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and M is the middle point of the line segment AB, then the position vector of M is \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\)).

Proof:

Let AB be a line segment and O be the origin.

Here,

Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)

Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)

Let M be the middle point of the segment AB.

Then,

\(\begin{align*} \overrightarrow {OM} &= \overrightarrow {OA} + \overrightarrow {AM}\\ &= \overrightarrow {OA} + \frac 12\overrightarrow {AB}\\ &= \overrightarrow {OA} + \frac 12(\overrightarrow {OB} - \overrightarrow {OA})\\ &= \overrightarrow a + \frac 12(\overrightarrow b - \overrightarrow a)\\ &= \frac {2\overrightarrow a + \overrightarrow b - \overrightarrow a}{2}\\ &= \frac 12(\overrightarrow a + \overrightarrow b)\\ \end{align*}\)

∴ Position Vector of M = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\))Proved

Theorem 2: (Section Formula for Internal Division)

If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and the point M divides the line segment AB internally in the ratio m : n, then the position vector of M is \(\frac {m \overrightarrow b + n\overrightarrow a}{m + n}\).

Proof:

Let AB be a line segment and O be the origin.

Here,

Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)

Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)

Let the point M divides AB internally in the ratio m : n.

Then,

\(\begin{align*} \overrightarrow {OM} &= \overrightarrow {OA} + \overrightarrow {AM}\\ &= \overrightarrow {OA} + \frac {m}{m + n}\overrightarrow {AB}\\ &= \overrightarrow {OA} + \frac m{m + n} (\overrightarrow {OB} - \overrightarrow {OA})\\ &= \overrightarrow a + \frac m{m + n} (-\overrightarrow a)\\ &= \frac {m\overrightarrow a + n\overrightarrow a + m\overrightarrow b - m\overrightarrow a}{m + n}\\ &= \frac {m\overrightarrow b + n\overrightarrow a}{m + n}\\ \end{align*}\)

∴ Position Vector of M = \(\frac {m\overrightarrow b + n\overrightarrow a}{m + n}\)Proved

Theorem 3: (Section Formula for External Division)

If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and the point P divides the line segment AB externally in the ratio m : n, then the position vector of P is \(\frac {m \overrightarrow b - n\overrightarrow a}{m - n}\).

Proof:

Let AB be a line segment and O be the origin.

Here,

Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)

Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)

Let the point P divides AB externally in the ratio m : n.

Then,

\(\frac {AP}{BP}\) = \(\frac mn\)

∴ n\(\overrightarrow {AP}\) = m\(\overrightarrow {BP}\)

or, n (\(\overrightarrow {OP}\) - \(\overrightarrow {OA}\)) = m(\(\overrightarrow {OP}\) - \(\overrightarrow {OB}\))

or, n\(\overrightarrow {OP}\) - n\(\overrightarrow {OA}\) = m\(\overrightarrow {OP}\) - m\(\overrightarrow {OB}\)

or, m\(\overrightarrow {OB}\) - n\(\overrightarrow {OA}\) = m\(\overrightarrow {OP}\) - n\(\overrightarrow {OP}\)

or, m\(\overrightarrow b\) - n\(\overrightarrow a\) = (m - n)\(\overrightarrow {OP}\)

∴ \(\overrightarrow {OP}\) = \(\frac {m\overrightarrow b - n\overrightarrow a}{m - n}\)

So, the position vector of P = \(\frac {m\overrightarrow b - n\overrightarrow a}{m - n}\) Proved

Theorem 4:

The line segment joining the mid-point of two sides of a triangle is parallel to the third side and it is half of it.



Proof:

Let ABC be a triangle and P and Q be the mid-points of the sides AB and AC respectively.

Here,

\(\overrightarrow {PA}\) = \(\frac 12\)\(\overrightarrow {BA}\)

\(\overrightarrow {AQ}\) = \(\frac 12\)\(\overrightarrow {AC}\)

Then,

\(\overrightarrow {PA}\) + \(\overrightarrow {AQ}\) = \(\frac 12\)\(\overrightarrow {BA}\) + \(\frac 12\)\(\overrightarrow {AC}\)

∴ \(\overrightarrow {PQ}\) = \(\frac 12\)(\(\overrightarrow {BA}\) + \(\overrightarrow {AC}\))

i.e. \(\overrightarrow {PQ}\) = \(\frac 12\)\(\overrightarrow {BC}\)

Clearly, \(\overrightarrow {PQ}\) // \(\overrightarrow {BC}\)Proved

Theorem 5:

The position vector of the centroid of a triangle is given by \(\overrightarrow g\) = \(\frac 13\) (\(\overrightarrow a\) + \(\overrightarrow b\) + \(\overrightarrow c\)) where \(\overrightarrow a\), \(\overrightarrow b\) and \(\overrightarrow c\) are the position vectors of the vertices and \(\overrightarrow g\) is the position vector of the centroid.

Proof:

Let ABC be triangle and O be the origin.

Let,

\(\overrightarrow {OA}\) = \(\overrightarrow a\)

\(\overrightarrow {OB}\) = \(\overrightarrow b\)

\(\overrightarrow {OC}\) = \(\overrightarrow c\)

Let D be the mid- point of AC.

By mid-point theorem,

\(\overrightarrow {OD}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow c\))

Let G be the centroid of the triangle ABC.

Then G divides BD internally in the ratio 2: 1.

Then,

\(\begin{align*} \overrightarrow {OG} &= \frac {2\overrightarrow {OD} + 1\overrightarrow {OB}}{2 + 1}\\ &= \frac {2 × \frac 12 \overrightarrow a + \overrightarrow c + \overrightarrow b}{3}\\ &= \frac 13 (\overrightarrow a + \overrightarrow b + \overrightarrow c)\\ \end{align*}\)

∴ Position Vector of G (\(\overrightarrow g\)) = \(\frac 13 (\overrightarrow a + \overrightarrow b + \overrightarrow c)\)Proved

Theorem 6:

The median to the base of an isosceles triangle is perpendicular to the base.

ABC be an isosceles triangle where AB = ACand AM is the median to the base BC.Proof:

Let \(\overrightarrow {AB}\) = \(\overrightarrow a\) and \(\overrightarrow {AC}\) = \(\overrightarrow b\)

Then,

|\(\overrightarrow a\)| = |\(\overrightarrow b\)| or a = b

Now,

\(\overrightarrow {AM}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BM}\) = \(\overrightarrow {AB}\) + \(\frac 12\)\(\overrightarrow {BC}\)

\(\overrightarrow {BC}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AC}\) = -\(\overrightarrow a\) + \(\overrightarrow b\)

\(\begin{align*} \therefore \overrightarrow {AM} &= \overrightarrow a + \frac 12 (\overrightarrow b - \overrightarrow a)\\ &= \frac {\overrightarrow a + \overrightarrow b - \overrightarrow c}2\\ &= \frac 12(\overrightarrow a + \overrightarrow b)\\ \end{align*}\)

Now,

\(\begin{align*} \overrightarrow {AM} . \overrightarrow {BC} &= \frac 12(\overrightarrow a + \overrightarrow b) . (\overrightarrow b - \overrightarrow a)\\ &= \frac 12 (\overrightarrow b + \overrightarrow a) . (\overrightarrow b - \overrightarrow a)\\ &= \frac 12(b^2 - a^2)\\ &= 0\\ \end{align*}\)

∴ AM⊥ BCProved

Theorem 7:

The figure formed by joining the mid-point of the adjacent sides of a quadrilateral is a parallelogram.

Proof:

Let ABCD be a quadrilateral and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Join BD.

Then,

\(\overrightarrow {BA}\) + \(\overrightarrow {AD}\) = \(\overrightarrow {BD}\)

2\(\overrightarrow {PA}\) + 2\(\overrightarrow {AS}\) = \(\overrightarrow {BD}\)

\(\overrightarrow {PA}\) + \(\overrightarrow {AS}\) = \(\frac 12\)\(\overrightarrow {BD}\)

∴ \(\overrightarrow {PS}\) = \(\frac 12\)\(\overrightarrow {BD}\)

Clearly, \(\overrightarrow {PS}\) // \(\overrightarrow {BD}\)

Again,

\(\overrightarrow {BC}\) + \(\overrightarrow {CD}\) = \(\overrightarrow {BD}\)

2\(\overrightarrow {QC}\) + 2\(\overrightarrow {CR}\) = \(\overrightarrow {BD}\)

\(\overrightarrow {QC}\) + \(\overrightarrow {CR}\) = \(\frac 12\)\(\overrightarrow {BD}\)

∴ \(\overrightarrow {QC}\) = \(\frac 12\)\(\overrightarrow {BD}\)

Clearly, \(\overrightarrow {QR}\) // \(\overrightarrow {BD}\)

\(\overrightarrow {PS}\) = \(\overrightarrow {QR}\), \(\overrightarrow {PS}\) // \(\overrightarrow {QR}\)

Similarly,

\(\overrightarrow {PQ}\) = \(\overrightarrow {SR}\), \(\overrightarrow {PQ}\) // \(\overrightarrow {SR}\)

Hence, PQRS is a parallelogram.

Theorem 8:

The diagonals of a parallelogram bisect each other.

Proof:

Let OACB be a parallelogram and O be the origin.

Let \(\overrightarrow {OA}\) = \(\overrightarrow a\) and \(\overrightarrow {OB}\) = \(\overrightarrow b\)

Let M be the middle point of AB.

Then,

\(\overrightarrow {OM}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\)).............................................(i)

Also,

\(\overrightarrow {OC}\) = \(\overrightarrow a\) + \(\overrightarrow b\), by parallelogram of vector addition.

Let N be the middle point of OC.

Then,

\(\overrightarrow {ON}\) = \(\frac 12\)\(\overrightarrow {OC}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\))............................................(ii)

From (i) and (ii), M and N have the same position vector. So, M and N are same points.

Hence, the diagonals of a parallelogram bisect each other.

Theorem 9:

The diagonals of a rectangle are equal.

Proof:

Let ABCD be a rectangle and AC and BD are diagonals.

Then,

\(\overrightarrow {AB}\) = \(\overrightarrow {DC}\), \(\overrightarrow {AD}\) = \(\overrightarrow {BC}\), \(\angle A\) = \(\angle B\) = \(\angle C\) = \(\angle D\) = 90° (\(\overrightarrow {AB}\) . \(\overrightarrow {BC}\) = 0, \(\overrightarrow {BA}\) . \(\overrightarrow {AD}\) = 0)

Now,

\(\overrightarrow {AC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BC}\)

\(\overrightarrow {AC}^2\) = (\(\overrightarrow {AB}\) + \(\overrightarrow {BC}\))2

AC2 = \(\overrightarrow {AB}^2\) + 2\(\overrightarrow {AB}\) . \(\overrightarrow {BC}\) + \(\overrightarrow {BC}^2\)

AC2= AB2 + 0 + BC2

AC2= AB2 + BC2......................................(i)

Again,

\(\overrightarrow {BD}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AD}\)

\(\overrightarrow {BD}^2\) = (\(\overrightarrow {BA}\) + \(\overrightarrow {AD}\))2

BD2= \(\overrightarrow {BA}^2\) + 2\(\overrightarrow {BA}\) . \(\overrightarrow {AD}\) + \(\overrightarrow {AD}^2\)

BD2= BA2 + 0 + AD2

BD2= AB2 + BC2......................................(i)

From (i) and (ii),

AC2= BD2

∴ AC = BD

Hence, the diagonals of a rectangle are equal.

Theorem 10:

The diagonals of a rhombus bisect each other at right angles.

Proof:

Let ABCD is a rhombus and AC and BD are diagonals.

Obviously,

\(\overrightarrow {AO}\) = \(\overrightarrow {OC}\) and \(\overrightarrow {BO}\) = \(\overrightarrow {OD}\)

Now,

We prove that,

\(\overrightarrow {AO}\)⊥ \(\overrightarrow {BO}\)

Let \(\overrightarrow {AB}\) = \(\overrightarrow a\) and \(\overrightarrow {AD}\) = \(\overrightarrow b\)

Then,

\(\overrightarrow {AC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {AD}\) = \(\overrightarrow a\) + \(\overrightarrow b\)

\(\overrightarrow {AO}\) = \(\frac 12\)\(\overrightarrow {AC}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\))..............................(i)

Again,

\(\overrightarrow {BD}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AD}\) = -\(\overrightarrow a\) + \(\overrightarrow b\) = \(\overrightarrow b\) - \(\overrightarrow a\)

\(\overrightarrow {BO}\) = \(\frac 12\)\(\overrightarrow {BD}\) = \(\frac 12\)(\(\overrightarrow b\) - \(\overrightarrow a\))..............................(ii)

From (ii) and (ii),

\(\overrightarrow {AO}\) .\(\overrightarrow {BO}\) =\(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\) .\(\frac 12\)(\(\overrightarrow b\) - \(\overrightarrow a\)) = \(\frac 14\) (\(\overrightarrow b^2\) - \(\overrightarrow a^2\)) = \(\frac 14\)(b2 - a2) = 0

Hence, the diagonals of a rhombus bisect each other at right angles.

Theorem 11:

The angle in a semi- circle is a right angle.

Proof:

Let O be the centre of a circle and AB be a diameter. \(\angle\)ACB is an angle in the semi- circle. Join OC.

Now,

\(\overrightarrow {AC}\) = \(\overrightarrow {AO}\) + \(\overrightarrow {OC}\)..............................(i)

\(\overrightarrow {CB}\) = \(\overrightarrow {CO}\) + \(\overrightarrow {OB}\) = \(\overrightarrow {CO}\) + \(\overrightarrow {AO}\)

\(\overrightarrow {CB}\) = \(\overrightarrow {AO} - \overrightarrow {OC}\)...................................(ii)

From (i) and (ii),

\(\begin{align*} \overrightarrow {AC} . \overrightarrow {CB} &= (\overrightarrow {AO} + \overrightarrow {OC}) . (\overrightarrow {AO} - \overrightarrow {OC})\\ & =(\overrightarrow {AO}^2 - \overrightarrow {OC}^2\\ &= AO^2 - OC^2 [since\; AO = OC]\\ &= 0\\ \end{align*}\)

Hence, the angle in the semi- circle is a right angle.

Theorem 12:

The mid- point of the hypotenuse of a right-angled triangle is equidistant from its vertices.

Proof:

Let AOC be a right angled triangle and O be the origin.

Let \(\angle AOC\) = 90° and M is the mid- point of AC.

Then,

\(\overrightarrow {OM}\) = \(\frac {\overrightarrow {OA} + \overrightarrow {OC}}2\)

2\(\overrightarrow {OM}\) = \(\overrightarrow {OA}\) + \(\overrightarrow {OC}\)

(2\(\overrightarrow {OM}\))2 = (\(\overrightarrow {OA}\) + \(\overrightarrow {OC}\))2

4 \(\overrightarrow {OM}^2\) = \(\overrightarrow {OA}^2\) + 2\(\overrightarrow {OA}\) . \(\overrightarrow {OC}\) + \(\overrightarrow {OC}^2\)

4 OM2 = OA2 + 0 + OC2

4 OM2 = OA2 + OC2

4 OM2 = AC2

OM2 = \(\frac {AC^2}4\)

∴ OM = \(\frac 12\)AC

Also,

AM = MC = \(\frac 12\)AC

∴ AM = MC = OM

Hence, the mid- point of the hypotenuse of a right angled triangle is equidistant from its vertices.

Dot product of \(\overrightarrow a\) & \(\overrightarrow b\)  = \(\overrightarrow a\) . \(\overrightarrow b\)
If \(\overrightarrow i\) and \(\overrightarrow j\) are unit vectors along x-axis and y-axis \(\overrightarrow i\).\(\overrightarrow i\) = \(\overrightarrow j\). \(\overrightarrow j\) = 1 and \(\overrightarrow i\).\(\overrightarrow j\) = \(\overrightarrow j\).\(\overrightarrow i\) = 0

Here,

\(\vec a\) = \(\begin {pmatrix} 6\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} -1\\ 6\\ \end {pmatrix}\)

\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 6× -1 + 1× 6 = 6 - 6 = 0

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(6)^2 + (1)^2}\) = \(\sqrt {36 + 1}\) = \(\sqrt {37}\)

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-1)^2 + (6)^2}\) = \(\sqrt {1 + 36}\) = \(\sqrt {37}\)

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)

or, cos\(\theta\) = \(\frac 0{\sqrt {37} \sqrt {37}}\)

or, cos\(\theta\) = 0

or, \(\theta\) = cos-1(0) = 90°

The angle between two vectors is 90° so, the \(\vec a\) and \(\vec b\) are perpendicular each other. Proved

Here,

\(\vec a\) = \(\begin {pmatrix} 2\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} 0\\ -2\\ \end {pmatrix}\)

\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 2 × 0 + 1 × -2 = 0 + -2 = -2

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (1)^2}\) = \(\sqrt {4 + 1}\) = \(\sqrt 5\)

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(0)^2 + (-2)^2}\) = \(\sqrt {0 + 4}\) = 2

cos\(\theta\) = \(\frac {-2}{2\sqrt 5}\) = -\(\frac 1{\sqrt 5}\)

∴ \(\theta\) = cos-1 (-\(\frac 1{\sqrt 5}\)) Ans

Here,

\(\vec p\) = \(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)

\(\vec p\) . \(\vec q\) = x1x2 + y1y2 = 3 × 1 + 4 × 1 = 3 + 4 = 7

\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt 25\) = 5

\(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(1)^2 + (1)^2}\) = \(\sqrt {1 + 1}\) = \(\sqrt 2\)

cos\(\theta\) = \(\frac {7}{5\sqrt 2}\) = \(\frac 7{7.07}\) = 0.99

∴ \(\theta\) = cos-1 (0.99) = 8.110Ans

Here,

\(\vec a\) = \(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\) and \(\vec c\) = \(\begin {pmatrix} 4\\ -2\\ \end {pmatrix}\)

\(\vec a\) . \(\vec c\) = x1x2 + y1y2 = 2× 4+ (-3) × (-2) = 8+ 6=14

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (-3)^2}\) = \(\sqrt {4 + 9}\) = \(\sqrt {13}\)

\(\begin {vmatrix} \vec c\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(4)^2 + (-2)^2}\) = \(\sqrt {16 + 4}\) = \(\sqrt {20}\)

cos\(\theta\) = \(\frac {14}{\sqrt {13} . \sqrt {20}}\) = \(\frac {14}{16.13}\) = 0.87

∴ \(\theta\) = cos-1 (0.87) = 29.540Ans

Here,

\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = 3, \(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = 3\(\sqrt 2\) and \(\vec p\).\(\vec q\) = 9

We know,

cos\(\theta\) = \(\frac {\vec p . \vec q}{\begin {vmatrix} \vec p\\ \end {vmatrix}\begin {vmatrix} \vec q\\ \end {vmatrix}}\) = \(\frac 9{3 × 3\sqrt 2}\) = \(\frac 1{\sqrt 2}\)

cos\(\theta\) = \(\frac 1{\sqrt 2}\)

\(\theta\) = cos-1(\(\frac 1{\sqrt 2}\)) = 45°

∴ The angle between \(\vec p\) and \(\vec q\) is 45°. Ans

Here,

\(\vec a\) = 4 \(\vec i\) + \(\vec j\) and \(\vec b\) = 3 \(\vec i\) + 4\(\vec j\)

(i)

\(\vec a\).\(\vec b\) = (4 \(\vec i\) + \(\vec j\)) . (3\(\vec i\) + 4\(\vec j\)) = 4× 3 + 1× 4 = 12 + 4 = 16 Ans

(ii)

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(4)^2 + (1)^2}\) = \(\sqrt {16 + 1}\) = \(\sqrt {17}\)

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5

Let: \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then,

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {16}{5\sqrt {17}}\)

\(\theta\) = cos-1(\(\frac {16}{5\sqrt {17}}\)) = 37.47°

∴ The angle between \(\vec a\) and \(\vec b\) is 37.47°. Ans

Here,

\(\vec a\) = -4\(\vec i\) + 5\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 10\(\vec j\)

(x1 , y1) = (-4 , 5) and (x2 , y2) = (8 , -10)

\(\vec a\).\(\vec b\) = x1x2 + y1y2 = -4 × 8 + 5 × 10 = -32 - 50 = -82

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-4)^2 + (5)^2}\) = \(\sqrt {16 + 25}\) = \(\sqrt {41}\)

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-10)^2}\) = \(\sqrt {64 + 100}\) = \(\sqrt {164}\)

We know,

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {-82}{\sqrt {41} \sqrt {164}}\) = - 1 (Approx.)

\(\theta\) = cos-1 (-1) = 180°

∴ The \(\vec a\) and \(\vec b\) are parallel in opposite direction. Proved

Here,

\(\vec a\) = 3 + k and \(\vec b\) = -7 + 3 are perpendiculat each other.

Then:

\(\vec a\) . \(\vec b\) = 0

or, (3 + k) (-7 + 3) = 0

or, 3× -7 + k× 3 = 0

or, -21 + 3k = 0

or, 3k = 21

or, k = \(\frac {21}3\)

∴ k = 7 Ans

Here,

\(\vec a\) = 4 + 2 and \(\vec b\) = -1 + 2

If \(\vec a\) and \(\vec b\) are perpendicular than \(\vec a\) . \(\vec b\) = 0

or, (4 + 2) (-1 + 2) = 0

or, 4× -1 + 2× 2 = 0

or, -4 + 4 = 0

∴ 0 = 0

∴ \(\vec a\) and \(\vec b\) are perpendicular to each other. Proved

Here,

\(\vec p\) = 10\(\vec i\) + 2k \(\vec j\) and \(\vec q\) = 2\(\vec i\) - 5\(\vec j\)

(x1, y1) = (10, 2k) and (x2, y2) = (2. -5)

We know,

\(\vec p\) and \(\vec q\) are perpendicular.

Then: \(\vec p\) . \(\vec q\) = 0

or, (10, 2k) (2, -5) = 0

or, 10× 2 + 2k× -5 = 0

or, 20 - 10k = 0

or, 10k = 20

or, k = \(\frac {20}{10}\)

∴ k = 2 Ans

Here,

\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)

\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 7 × 5 + (-5) × (-7) = 35 + 35 = 70

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(7)^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(5)^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) =\(\frac {70}{\sqrt {74} \sqrt {74}}\) = \(\frac {70}{74}\) = 0.95

\(\theta\) = cos-1(0.95) = 18.93°

∴\(\angle\)AOB = 18.93° Ans

Let: the vector \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar quantity.

\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and b = 8\(\vec i\) - 6\(\vec j\)

\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 3 × 8 + 4 × (-6) = 24 - 24 = 0

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 × 10}\) = 0

∴ \(\theta\) = cos-1 (0) = 90° Ans

Here,

\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?

Using triangle law in vector addition,

\(\vec {OA}\) + \(\vec {AB}\) = \(\vec {OB}\)

\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)

We know,

\(\vec {AC}\) = 2\(\vec {AB}\) = 3 (\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)

Again,

\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)

∴ \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\) = 3\(\vec b\) - 2\(\vec a\) Ans

Given points are: A(1, 2) and B(3, 0)

x-component = x2 - x1 = 3 - 1 = 2

y-component = y2 - y1 = 0 - 2 = - 2

\(\vec {AB}\) = (2, -2)

\(\vec {AB}\) = 2\(\vec i\) - 2\(\vec j\) Ans

Given points are: C(1, 1) and D(-2, -4)

x-component = x2 - x1 = -2 - 1 = -3

y-component = y2 - y1 = 0 - 2 = - 2

\(\vec {CD}\) = (-3, -5)

\(\vec {CD}\) = -3\(\vec i\) - 5\(\vec j\) Ans

Given points are: P(3, 5) and Q(-7, 3).

Let, O be the origin.

\(\vec {OP}\) = \(\begin {pmatrix} 3\\ 5\\ \end {pmatrix}\) and \(\vec {OQ}\) = \(\begin {pmatrix} -7\\ 3\\ \end {pmatrix}\)

Using midpoint formula,

\(\vec {OM}\) = \(\frac {\vec {OP} + \vec {OQ}}2\)

or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3\\ 5\\ \end{pmatrix} + \begin {pmatrix} -7\\ 3\\ \end {pmatrix}}2\)

or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3 - 7\\ 5 + 3\\ \end {pmatrix}}2\)

or, \(\vec {OM}\) = \(\frac {\begin {pmatrix} -4\\ 3\\ \end {pmatrix}}2\)

or, \(\vec {OM}\) = \(\begin {pmatrix} \frac {-4}2\\ \frac 82\\ \end {pmatrix}\)

∴ \(\vec {OM}\) = \(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)

∴ \(\vec {OM}\) = -2\(\vec i\) + 4\(\vec j\) Ans

Here,

D is the midpoint of the line BC of the \(\triangle\)ABC.

In \(\triangle\)ABD,

\(\vec {AB}\) = \(\vec {AD}\) + \(\vec {DB}\)....................(1) [\(\because\) triangle law of vector addition]

In \(\triangle\)ACD,

\(\vec {AC}\) = \(\vec {AD}\) + \(\vec {DC}\)...................(2)[\(\because\) triangle law of vector addition]

Adding (1) and (2)

\(\vec {AB}\) + \(\vec {AC}\) = \(\vec {AD}\) + \(\vec {BD}\) + \(\vec {AD}\) + \(\vec {DC}\)

or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + (\(\vec {DB}\) + \(\vec {DC}\))

or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + \(\vec {DB}\) - \(\vec {DB}\) [\(\because\) -\(\vec {DB}\) = \(\vec {DC}\)]

∴ \(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) Ans

Let, O be the origin.

\(\vec {OA}\) = 3\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 6\(\vec j\)

We know,

\(\vec {OM}\) = \(\frac {\vec {OA} + \vec {OB}}2\)

or, \(\vec {OM}\) = \(\frac {3\vec i + 2\vec j + 5\vec i - 6\vec j}2\)

or, \(\vec {OM}\) = \(\frac {8\vec i - 4\vec j}2\)

∴ \(\vec {OM}\) = 4\(\vec i\) - 2\(\vec j\)

∴ The position vector of M = 4\(\vec i\) - 2\(\vec j\) Ans

Let: O be the origin.

\(\vec {OA}\) = 5\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 3\(\vec i\) + 6\(\vec j\)

m : n = 2 : 3

Using section internal division formula.

\(\vec {OP}\) = \(\frac {m \vec{OB} + n\vec {OA}}{m + n}\)

or, \(\vec {OP}\) = \(\frac {2(3\vec i + 6\vec j) + 3(5\vec i + 2\vec j)}{2 + 3}\)

or, \(\vec {OP}\) = \(\frac {6\vec i + 12\vec j + 15\vec i + 6\vec j}5\)

or, \(\vec {OP}\) = \(\frac {21\vec i + 18\vec j}5\)

∴ \(\vec {OP}\) = \(\frac {21}5\)\(\vec i\) + \(\frac {18}5\)\(\vec j\) Ans

Here,

A(-4, 8) and B(3, 7)

Let, O be the origin.

\(\vec {OA}\) = \(\begin {pmatrix} -4\\ 8\\ \end {pmatrix}\)

\(\vec {OB}\) = \(\begin {pmatrix} 3\\ 7\\ \end {pmatrix}\)

Using section external division formula,

\(\vec {AB}\) = \(\frac {4\begin {pmatrix} 3\\ 7\\ \end {pmatrix} - 3\begin {pmatrix} -4\\ 8\\ \end {pmatrix}}{4 - 3}\)

or, \(\vec {AB}\) = \(\frac {\begin {pmatrix} 12\\ 28\\ \end {pmatrix} - \begin {pmatrix} -12\\ 24\\ \end {pmatrix}}1\)

or, \(\vec {AB}\) = \(\begin {pmatrix} 12 + 12\\ 28 - 24\\ \end {pmatrix}\)

or, \(\vec {AB}\) = \(\begin {pmatrix} 24\\ 4\\ \end {pmatrix}\)

∴ \(\vec {AB}\) = 24\(\vec i\) + 4\(\vec j\) Ans

Here,

\(\vec {OA}\) = \(\begin {pmatrix} \sqrt 3\\ 1\\ \end {pmatrix}\) and \(\vec {OB}\) = \(\begin {pmatrix} \sqrt 3\\ 3\sqrt 3\\ \end {pmatrix}\)

Let: x1 = \(\sqrt 3\), y1 = 1, x2 = \(\sqrt 3\) and y2 = 3\(\sqrt 3\)

Let: \(\theta\) be the angle between \(\vec {OA}\) and \(\vec {OB}\)

cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)

or, cos \(\angle\)AOB = \(\frac {\sqrt 3 × \sqrt 3 + 1 × 3\sqrt 3}{\sqrt {(\sqrt 3)^2 + 1^2} \sqrt {(\sqrt 3)^2 + (3\sqrt 3)^2}}\)

or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{\sqrt {3 + 1}. \sqrt {3 + 27}}\)

or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{2\sqrt {30}}\)

or, cos \(\angle\)AOB = \(\frac {8.197}{10.95}\)

or, cos \(\angle\)AOB = 0.75

or, \(\angle\)AOB = cos-1 (0.75)

∴\(\angle\)AOB = 41.58° Ans

DF

If the vector is \(\vec u\) and \(\vec v\) are represented by the two adjacent sides \(\vec {AB}\) and \(\vec {AD}\) of a parallelogram ABCD, then the sum \(\vec u\) + \(\vec v\) is represented by the diagonal \(\vec {AC}\).

\(\vec {AB}\) + \(\vec {AD}\) = \(\vec {AC}\)

\(\vec u\) + \(\vec v\) = \(\vec {AC}\)

Unit vector: A vector whose modulus is 1 is called a unit vector.

\(\vec a\) = (\(\frac 1{\sqrt 2}\), \(\frac 1{\sqrt 2}\)) is unit vector.

\(\vec a\) . \(\vec b\)

= (4\(\vec i\) - 6\(\vec j\)) . (3\(\vec i\) + 2\(\vec j\))

= 12\(\vec {i^2}\) + 8\(\vec i\) \(\vec j\) - 18\(\vec i\) \(\vec j\) - 12\(\vec {j^2}\)

= 12× 1 + 8× 0 - 18× 0 - 12× 1 [\(\vec {i^2}\) = \(\vec {j^2}\) = 1, \(\vec i\) . \(\vec j\) = 0]

= 12 - 12

= 0

\(\vec a\) . \(\vec b\) = 0

∴ \(\vec a\) . \(\vec b\) are perpendicular each other. Proved

Here,

\(\vec a\) = (4\(\vec i\) + 5\(\vec j\)) and \(\vec b\) = (5\(\vec i\) - 4\(\vec j\))

\(\vec a\) = (4, 5) and \(\vec b\) = (5, -4)

If \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then:

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)

or, cos\(\theta\) = \(\frac {(4, 5) . (5, -4)}{\sqrt {4^2 + 5^2} \sqrt {5^2 + (-4)^2}}\)

or, cos\(\theta\) = \(\frac {20 - 20}{\sqrt {16 + 25} \sqrt {25 + 16}}\)

or, cos\(\theta\) = \(\frac 0{25 + 16}\)

or, cos\(\theta\) = 0

or, \(\theta\) = cos-1 (0)

∴ \(\theta\) = 90° Ans

Here,

\(\vec a\) = 10\(\vec i\) - 7\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 10\(\vec j\)

If \(\theta\) be the angle between the vectors \(\vec a\) and \(\vec b\).

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)

or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)

or, cos\(\theta\) = \(\frac {10 × 7 + 10 × (-7)}{\sqrt {(10)^2 + (-7)^2} \sqrt {7^2 + {10}^2}}\)

or, cos\(\theta\) = \(\frac {70 - 70}{\sqrt {149} \sqrt {149}}\)

or, cos\(\theta\) = 0°

∴ \(\theta\) = cos-1(0°) = 90° Ans

In the \(\triangle\)ABC,

Let O be the origin.

Position vector of the point A is = \(\vec {OA}\) = (-1, -1)

Position vector of the point B is = \(\vec {OB}\) = (5, -1)

Position vector of the point C is = \(\vec {OC}\) = (2, 5)

If G be the centroid of the triangle,

= \(\frac 13\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\))

= \(\frac 13\) (- 1 - 1 + 5 - 1 + 2 + 5)

= \(\frac 13\) (6 + 3)

= 2 + 1

Hence, the position vector of the centroid is (2, 1). Ans

Here,

\(\vec a\) = 6\(\vec i\) - 8\(\vec j\) and \(\vec b\) = 4\(\vec i\) + 3\(\vec j\)

\(\vec a\) . \(\vec b\)

= (6\(\vec i\) - 8\(\vec j\)) (4\(\vec i\) + 3\(\vec j\))

= 6 . 4\(\vec {i^2}\) + 18\(\vec i\) \(\vec j\) - 32 \(\vec i\) \(\vec j\) - 24\(\vec {j^2}\)

= 24 . 1 + 18 . 0 - 32 . 0 * 24 . 1 [\(\because\) \(\vec {i^2}\) = \(\vec {j^2}\) = 1 , \(\vec j\) = \(\vec i\) \(\vec j\) = 0]

= 24 - 24

= 0

∴ The dot product of two vector a and b is equal to zero so these are perpendicular to each other. Proved

Here,

\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)

\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 7× 5 + (-5) (-7) = 35 + 35 = 70

\(\begin {vmatrix} \vec a \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {7^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)

\(\begin {vmatrix} \vec b\end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {5^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b\end {vmatrix}}\)

or, cos\(\theta\) = \(\frac {70}{\sqrt {74} . \sqrt {74}}\)

or, cos\(\theta\) = \(\frac {70}{74}\)

or, cos\(\theta\) = 0.95

or, \(\theta\) = cos-1 (0.95)

∴ \(\theta\) = 18.93°

∴ \(\angle\)AOB = 18.93° Ans

Unit Vector: If the magnitude of the vector is 1 such type of vector is known as unit vector.

i.e., \(\vec {OP}\) = \(\begin {pmatrix} 1\\ 0\\ \end {pmatrix}\)

Let, \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 8\(\vec j\) are the position vectors of A and B,

Position vector of mid-point \(\vec M\) = \(\frac 12\) (\(\vec a\) + \(\vec b\))

\(\vec M\) = \(\frac 12\) (3\(\vec i\) + 4\(\vec j\) + 7\(\vec i\) + 8\(\vec j\)) = \(\frac 12\) (10\(\vec i\) + 12\(\vec j\)) = 5\(\vec i\) + 6\(\vec j\) Ans

Here,

\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and\(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?

Using triangle law of vector addition:

\(\vec {OA}\) + \(\vec {AB}\) \(\vec {OB}\)

or, \(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)

We know,

\(\vec {AC}\) = 3\(\vec {AB}\) = 3(\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)

Again,

\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)

or, \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\)

∴ \(\vec {OC}\) = 3\(\vec b\) - 2\(\vec a\) Ans

The two vectors are \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar product.

\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\)

\(\vec a\).\(\vec b\) = x1x2 + y1y2 = 3× 8 + (-4) × 6= 24 - 24 = 0

\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5

\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10

We know,

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 × 10}\) = 0

∴ \(\theta\) = cos-1 (0) = 90° Ans

Let: \(\vec a\) = \(\vec i\) - 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) - 5\(\vec j\) , m:n = 3:1

Position vector of C (\(\vec {OC}\)) = \(\frac {n\vec a + m \vec b}{m + n}\)

\(\vec c\) = \(\vec {1 (\vec i - 3\vec j)}{3 + 1}\)

\(\vec c\) = \(\frac {\vec i - 3\vec j + 6\vec i + 15\vec j}{4}\)

\(\vec c\) = \(\frac {7\vec i + 12\vec j}4\) Ans

\(\vec {AB}\) = \(\begin {pmatrix} x_2 - x_1\\ y_2 - y_1\\ \end {pmatrix}\) =\(\begin {pmatrix} 2 - 1\\ 5 + 3\\ \end {pmatrix}\) = \(\begin {pmatrix} 1\\ 8\\ \end {pmatrix}\)

\(\begin {vmatrix} \vec {AB}\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {1^2 + 8^2}\) = \(\sqrt {1 + 64}\) = \(\sqrt {65}\) Ans

Here,

Using triangle law of vector addition,

\(\vec {OC}\) = \(\vec {OA}\) + \(\vec {AC}\)

or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec {AB}\)

or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec {OB}\) - \(\vec {OA}\))

or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec b\) - \(\vec a\))

or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec b\) - 5\(\vec a\)

∴ \(\vec {OC}\) = 5\(\vec b\) - 4\(\vec a\) Ans

Orthogonal Vectors: If two vectors are perpendicular each other then vector are known as orthogonal vector.

\(\vec p\) . \(\vec q\) = 3× 2 - 2× 3 = 0

∴ The dot product od \(\vec p\) and \(\vec q\) is equal to zero, so these vectors are orthogonal. Proved

Let: \(\vec {OA}\) = 3\(\vec i\) - 2\(\vec j\) and \(\vec {OB}\) = 1\(\vec i\) + 8\(\vec j\)

Mid-point of \(\vec {AB}\)

= \(\frac {\vec {OA} + \vec {OB}}2\)

= \(\frac {3\vec i - 2\vec j + 1\vec i + 8\vec j}2\)

= \(\frac {4\vec i + 6\vec j}2\)

= 2\(\vec i\) + 3\(\vec j\)

∴ Position vector of the mid-point of \(\vec {AB}\) = 2\(\vec i\) + 3\(\vec j\) Ans

czv

Position Vectors: Let P(x, y) be any point on the co-ordinates axes, then (x, y) is called the position vector of the point P with referred to the origin O. OP is a position vector.

Let: \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = -\(\vec i\) + 2\(\vec j\)

\(\vec m\) = ?

\(\vec m\) = \(\frac {\vec a + \vec b}2\) = \(\frac {3\vec i + 4\vec j - \vec i + 2\vec j}2\) = \(\frac {2\vec i + 6\vec j}2\) = \(\vec i\) + 3\(\vec j\) Ans

ASD

Using triangle law of vector addition,

In \(\triangle\)QRS,

\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) ....................................(1)

In \(\triangle\)PQS,

\(\vec {QS}\) = \(\vec {QP}\) + \(\vec {PS}\) ...................................(2)

From (1) and (2),

\(\vec {QR}\) + \(\vec {RS}\) = \(\vec {QP}\) + \(\vec {PS}\)

or, \(\vec {QR}\) + \(\vec {RS}\) - \(\vec {PS}\) = \(\vec {QP}\)

∴ \(\vec {QR}\) + \(\vec {RS}\) +\(\vec {SP}\) = \(\vec {QP}\) Proved

Given,

\(\vec a\) = \(\vec i\) + 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) + \(\vec j\)

Let an anglebetween \(\vec a\) and \(\vec b\) be \(\theta\)

So,

cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b \end {vmatrix}}\)

or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {{x_1}^2 +{y_1}^2} \sqrt {{x_2}^2 + {y_2}^2}}\) [where x1 = 1, y1 = 3, x2 = 2 and y2 = 1]

or, cos\(\theta\) = \(\frac {1 × 2 + 3 × 1}{\sqrt {1^2 + 3^2} \sqrt {2^2 + 1^2}}\)

or, cos\(\theta\) = \(\frac {2 + 3}{\sqrt {10} \sqrt 5}\)

or, cos\(\theta\) = \(\frac 5{\sqrt {50}}\)

or, cos\(\theta\) = \(\frac 5{5\sqrt 2}\)

or, cos\(\theta\) = \(\frac 1{\sqrt 2}\)

or, cos\(\theta\) = cos 45°

∴ \(\theta\) = 45° Ans

Null Vector: If the magnitude of vector is equal to zero such type of vector is known as zero or null vector. \(\vec {AA}\) = 0

If \(\vec a\) and \(\vec b\) are perpendicular then,

\(\vec a\) . \(\vec b) = 0

or, (-5, 3) (p, p+2) = 0

or, -5× p + 3× (p + 2) = 0

or, -5p + 3p + 6 = 0

or, -2p = - 6

or, p = \(\frac 62\)

∴ p = 3 Ans

Joining point P and R,

In \(\triangle\)RSP,

\(\vec {RP}\) = \(\vec {RS}\) + \(\vec {SP}\).................................(1)

In \(\triangle\)PQR,

\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)..............................(2)

Adding equation (1) and (2),

\(\vec {RP}\) + \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)

or, -\(\vec {PR}\)+ \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) \(\vec {RP}\) = -\(\vec {PR}\)]

or, 0= \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)

∴ \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) = 0 Proved

\(\vec {AB}\), \(\vec {BC}\) and \(\vec {AC}\) are the vertices which represent the side of \(\triangle\)ABC using triangle law of addition.

\(\vec {BC}\) + \(\vec {CA}\) = \(\vec {BA}\)

or, \(\vec {BC}\) + \(\vec {CA}\) = -\(\vec {AB}\) [\(\because\) \(\vec {BA}\) = - \(\vec {AB}\)]

∴ \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) = 0 Proved

In \(\triangle\)ABC,

\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)............................(1)

In \(\triangle\)ACD,

\(\vec {AC}\) + \(\vec {CD}\) = \(\vec {AD}\)...........................(2)

In \(\triangle\)ADE,

\(\vec {AD}\) + \(\vec {DE}\) = \(\vec {AE}\).............................(3)

or, \(\vec {AC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)

or, \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)

∴\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) + \(\vec {EA}\) = 0 Proved

Here,

Using triangle law of vector addition,

In \(\triangle\)OBC,

\(\vec {BC}\) = \(\vec {BO}\) + \(\vec {OC}\)

∴ \(\vec {BC}\) = -\(\vec b\) + \(\vec c\)

In \(\triangle\)OAD,

\(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\) = \(\vec {OA}\) + \(\vec {BC}\) [\(\vec {AD}\) = \(\vec {BC}\)]

∴ \(\vec {OD}\) = \(\vec a\) - \(\vec b\) + \(\vec c\) Ans

FG

Let, PQRS be a parallelogram which PR and SQ are equal diagonals.

Let: \(\vec {SR}\) = \(\vec a\) and \(\vec {SP}\) = \(\vec b\)

To prove: PQRS is a parallelogram.

\(\vec {SQ}\) = \(\vec {SR}\) + \(\vec {RQ}\) [\(\because\) Triangle law of vector addition]

\(vec {SR}\) = \(\vec {SR} \) + \(\vec {SP}\) [\(\because\) \(\vec {RQ}\) = \(\vec {SP}\)]

\(\vec {SQ}\) = \(\vec a\) + \(\vec b\)

\(\vec {RP}\) =\(\vec {RS}\) + \(\vec {SP}\) = -\(\vec {SR}\) + \(\vec {SP}\) = -\(\vec a\) \(\vec b\) = \(\vec b\) - \(\vec a\)

\(\begin {vmatrix} \vec {SQ}\\ \end {vmatrix}\) = \(\begin {vmatrix} \vec {RP}\\ \end {vmatrix}\)

or, \(\vec {SQ}\)2 = \(\vec {RP}\)2

or, (\(\vec a\) + \(\vec b\))2 = (\(\vec b\) - \(\vec a\))2

or, \(\vec a\)2 + 2\(\vec a\) \(\vec b\) + \(\vec b\)2 = \(\vec b\)2- 2\(\vec a\) \(\vec b\) + \(\vec a\)2

or, 4\(\vec a\) \(\vec b\) = 0

or, \(\vec a\) . \(\vec b\) = 0

∴ \(\angle\)PSR = 90°

Hence, PQRS is a rectangle. Proved

XCV

Let: ABCD is a rhombus.

Let: \(\vec {AB}\) = \(\vec a\) and \(\vec {AD}\) \ \(\vec b\)

To prove: AC ⊥BD and AO = \(\frac 12\) AC, BO = \(\frac {BD}\).

In \(\triangle\)ABC,

\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\) [\(\because\) Triangle law of vector addition]

\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {AD}\) [\(\because\) BC = AD]

\(\vec {AC}\) = \(\vec a\) + \(\vec b\)

\(\vec {BD}\) = \(\vec {BA}\) + \(\vec {AD}\) = -\(\vec a\) + \(\vec b\) = \(\vec b\) - \(\vec a\)

\(\vec {AC}\) . \(\vec {BD}\)

= (\(\vec a\) + \(\vec b\)) (\(\vec b\) - \(\vec a\))

= \(\vec b\)2 - \(\vec a\)2

=\(\vec {AD}\)2 - \(\vec {AB}\)2

= \(\vec {AB}\)2 - \(\vec {AB}\)2 [\(\because\) AB = AD]

= 0

∴ AC ⊥ BD

\(\vec {AO}\)= \(\vec {AB}\) + \(\vec {BO}\) = \(\vec {AB}\) + \(\frac 12\) \(\vec {BD}\) = a + \(\frac 12\)(\(\vec b\) - \(\vec a\))

\(\vec {AO}\) = \(\frac {2a + \vec b - \vec a}2\) = \(\frac 12\)(\(\vec a\) + \(\vec b\)) = \(\frac 12\)\(\vec {AC}\)..........................(1)

\(\vec {BO}\) = \(\vec {BA}\) + \(\vec {AO}\) = -\(\vec {AB}\) + \(\vec {AO}\) = - \(\vec a\) + \(\frac 12\)(\(\vec a\) + \(\vec b\)

\(\vec {BO}\) = \(\frac {-2\vec a + \vec a + \vec b}2\) = \(\frac 12\)(\(\vec b\) - \(\vec a\)) = \(\frac 12\)\(\vec {BD}\)............(2)

From relation (1) and (2):

Diagonal bisect each other.

Hence, diagonals of a rhombus bisects each at right angle. Proved

FV

Let: PQRS is a parallelogram, in which PQ//SR and PS//QR.

To prove: PQ = SR and PS = QR

Let \(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\) where m and n are scalars.

In \(\triangle\)PQR,

\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)

\(\vec {PR}\) = m\(\vec {SR}\) + n\(\vec {QR}\)......................(1)

In \(\triangle\)PSR,

\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\)...............................(2)

From (1) and (2),

m\(\vec {SR}\) + n\(\vec {QR}\) = \(\vec {PS}\) + \(\vec {SR}\)

or, m\(\vec {SR}\) - \(\vec {SR}\) = \(\vec {QR}\) - n\(\vec {QR}\)

or, \(\vec {SR}\) (m - 1) = \(\vec {QR}\) (1 - n)

\(\vec {QR}\) and \(\vec {SR}\) are not parallel so;

m - 1 = 0∴ m = 1

1 - n = 0∴ n = 1

Putting the value of m and n in

\(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\)

\(\vec {PQ}\) = \(\vec {SR}\) and \(\vec {PS}\) = \(\vec {QR}\)

Hence, opposite sides of a parallelogram are equal. Proved

DS

ABCD is a parallelogram, diagonals AC and BD are perpendicular to each other.

To prove: ABCD is a rhombus.

Let: \(\vec {AB}\) = \(\vec {DC}\) = \(\vec a\) and \(\vec {BC}\) = \(\vec {AD}\) = \(\vec b\)

Using triangle law of vector addition:

In \(\triangle\)ABC,

\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\)

\(\vec {AC}\) = \(\vec a\) + \(\vec b\)

In \(\triangle\)BCD,

\(\vec {BD}\) = \(\vec {BC}\) + \(\vec {CD}\)

\(\vec {BD}\) = \(\vec {BC}\) - \(\vec {DC}\) = \(\vec b\) - \(\vec a\)

AC and BD are perpendicular to each other.

\(\vec {AC}\) . \(\vec {BD}\) = 0

(\(\vec a\) + \(\vec b\)) + (\(\vec b\) - \(\vec a\)) = 0

or, \(\vec {b^2}\) - \(\vec {a^2}\) = 0

or, \(\vec {b^2}\) = \(\vec {a^2}\)

or, \(\begin {vmatrix} \vec {b^2} \end {vmatrix}\) =\(\begin {vmatrix} \vec {a^2} \end {vmatrix}\)

∴\(\begin {vmatrix} \vec b\end {vmatrix}\) =\(\begin {vmatrix} \vec a\end {vmatrix}\)

AB = BC

Hence, ABCD is a rhombus. Proved

FDG

PQRS is a parallelogram, PR and QS are diagonals.

To prove: PR2+ QS2 = PQ2 + QR2 + RS2 + PS2

In \(\triangle\)PQR,

\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) Triangle law of vector addition]

Squaring on both sides,

\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))2

\(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)

PR2 = PQ2 + QR2 + 2 PQ. QR.....................(1)

In \(\triangle\)QRS,

\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) [\(\because\) Triangle law of vector addition]

Squaring on both sides,

\(\vec {QS}^2\) = (\(\vec {QR}\) + \(\vec {RS}\))2 = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {RS}\) + \(\vec {RS}^2\)

\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {QP}\) + \(\vec {RS}^2\) [\(\because\) \(\vec {RS}\) = \(\vec {QP}\)]

\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . (-\(\vec {PQ}\)) + \(\vec {RS}^2\)

\(\vec {QS}^2\) = \(\vec {QR}^2\) -2 . \(\vec {QR}\) . \(\vec {PQ}\) + \(\vec {RS}^2\)

QS2 = QR2 + RS2 - 2 . PQ . QR............................(2)

Adding (1) and (2)

PR2 + QS2 = PQ2 + QR2 + QR2 + RS2 + 2 PQ . QR - 2 PQ . QR

∴PR2 + QS2 = PQ2 + QR2 + PS2 + RS2Hence, Proved

PQRS is a trapezium in which PQ\\SR. A and B are the mid-points of PR and QS. Let, P be the origin.

To prove: i. AB//PQ//SR ii. AB = \(\frac 12\)(PQ - SR)

Construction: join PB

Let, P be the origin.

\(\vec {SR}\) = m\(\vec {PQ}\) [\(\because\) PQ//SR]

\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\) [Using triangle law of vector addition]

\(\vec {PR}\) = \(\vec {PS}\) + m\(\vec {PQ}\)

\(\vec {PB}\) = \(\frac {\vec {PS} + \vec {PQ}}2\) [\(\because\) B is the mid-point of QS]

\(\vec {PA}\) = \(\frac 12\)\(\vec {PR}\) = \(\frac 12\)(\(\vec {PS}\) + m\(\vec {PQ}\))

\(\vec {AB}\) = \(\vec {PB}\) - \(\vec {PA}\) = \(\frac 12\)(1 - m)\(\vec {PQ}\)

\(\vec {AB}\)//\(\vec {PQ}\)

∴ AB//PQ//SR

Again,

\(\vec {AB}\) = \(\frac 12\)\(\vec {PQ}\) - \(\frac 12\)m\(\vec {PQ}\) = \(\frac 12\)(\(\vec {PQ}\) - m\(\vec {PQ}\)) = \(\frac 12\)(\(\vec {PQ}\) - \(\vec {SR}\))

∴ AB = \(\frac 12\)(PQ - SR) Proved

Given:

PQRS is a parallelogram.

In which: PQ =SR, PQ//SR and PS = QR, PS//QR

To prove: PMRN is a parallelogram.

In \(\triangle\)PQN,

\(\vec {PQ}\) = \(\vec {PN}\) + \(\vec {NQ}\)....................(1) [Using triangle law of vector addition]

In \(\triangle\)SMR,

\(\vec {SR}\) = \(\vec {SM}\) + \(\vec {MR}\).....................(2)

From equation (1) and (2)

\(\vec {PQ}\) = \(\vec {SR}\)

or, \(\vec {PN}\) + \(\vec {NQ}\) = \(\vec {SM}\) + \(\vec {MR}\)

\(\vec {PN}\) = \(\vec {MR}\) [\(\because\) \(\vec {NQ}\) = \(\vec {SM}\)]

PN and MR are equal and parallel.

Similarly, PM and NR are equal and parallel.

∴ PMRN is a parallelogram. Proved

AS

In the figure, ABC is a triangle with medians. AE, BF and CD of sides BC, AC and AB respectively. BE = \(\frac 12\)BC, CF = \(\frac 12\)CA, AD = \(\frac 12\)AB.

To prove:\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) = \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)

Using triangle law of vector addition,

\(\vec {AE}\) = \(\vec {AB}\) + \(\vec {BE}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)........................(1)

\(\vec {BF}\) = \(\vec {BC}\) + \(\vec {CF}\) = \(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\)....................(2)

\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)......................(3)

Adding (1), (2) and (3),

\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\)

= \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\) +\(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\) +\(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)

= \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\))

= \(\vec {AC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AC}\) + \(\vec {CA}\)) [\(\because\) \(\vec {AB}\) + \(\vec{BC}\) = \(\vec {AC}\)]

\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\) = -\(\vec {CA}\) + \(\vec {CA}\) + \(\frac 12\)(-\(\vec {CA}\) + \(\vec {CA}\)) = 0

or, \(\vec {AO}\) + \(\vec {OE}\) + \(\vec {BO}\) + \(\vec {OF}\) + \(\vec {CO}\) + \(\vec {OD}\) = 0 [\(\because\) \(\vec {AE}\) = \(\vec {AO}\) + \(\vec {OE}\)]

or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) - \(\vec {OA}\) - \(\vec {OB}\) - \(\vec {OC}\) = 0

or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) = \(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\)

∴\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) =\(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) Hence, Proved

Given:

ABCD is a trapezium.

AD//BC//MN and M and N are the mid-point of AB and DC.

To prove:MN = \(\frac 12\)(AD + BC)

Construction: join point A and N, join point B and N

Using triangle law of vector addition:

In \(\triangle\)BMN,

\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BN}\) = \(\vec {MB}\) + (\(\vec {BC}\) + \(\vec {CN}\))........................(1)

In \(\triangle\)MAN,

\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\) = \(\vec {MB}\) + (\(\vec {AD}\) + \(\vec {DN}\))........................(2)

Adding (1) and (2)

2\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {CN}\) + \(\vec {MA}\) + \(\vec {AD}\) + \(\vec {DN}\)

or, 2\(\vec {MN}\) = (\(\vec {MB}\) + \(\vec {BM}\)) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)

or, 2\(\vec {MN}\) = \(\vec {MB}\) - \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)

or, 2\(\vec {MN}\) = \(\vec {BC}\) + \(\vec {AD}\)

∴MN = \(\frac 12\)(AD + BC) Hence, Proved

dsf

Given:

In \(\triangle\)ABC, M and N are the mid-point of Ab and AC. M and N are joined.

To prove: \(\vec {MN}\) //\(\vec {BC}\) and \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)

Using triangle law of vector addition,

In \(\triangle\)AMN,

\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\).........................(1)

In \(\triangle\)ABC,

\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\)...........................(2)

Since, M and N are mid-point of BA and AC.

\(\vec {BC}\) = 2\(\vec {MA}\) + 2\(\vec {AN}\)

\(\vec {BC}\) = 2 (\(\vec {MA}\) + \(\vec {AN}\))

\(\frac {\vec {BC}}2\) = \(\vec {MN}\) [From equation (1)]

\(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)

\(\vec {MN}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) where, m is a scalar quantity then \(\vec a\)//\(\vec b\)]

∴ \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\) and \(\vec {MN}\)//\(\vec {BC}\) Hence, Proved

Given:

In \(\triangle\)ABC, P and Q are the mid points of AB and AC.

To Prove: \(\vec {PQ}\)// \(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)

Using triangle law of vector addition:

In \(\triangle\)APQ,

\(\vec {PQ}\) = \(\vec {PA}\) + \(\vec {AQ}\)....................................(1)

In \(\triangle\)ABC,

\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\).....................................(2)

Dividing by 2 on both sides of equation (2)

\(\frac 12\)\(\vec {BC}\) =\(\frac 12\)(\(\vec {BA}\) + \(\vec {AC}\))

\(\frac 12\)\(\vec {BC}\) = \(\vec {PA}\) + \(\vec {AQ}\) [\(\because\) P and Q are the mid-point of AB and AC]

\(\frac 12\)\(\vec {BC}\) = \(\vec {PQ}\) [From equation (1)]

∴ \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)

\(\vec {PQ}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]

Hence, \(\vec {PQ}\)//\(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\) Hence, Proved

sdf

AC and BD are the diagonals of a parallelogram of ABCD which intersect at a point G.GA = GC and GB = GD. If O be any point, then join O and A, O and B, O and C, O and D and O and G.

Using triangle law in vector addition,

In \(\triangle\)OAC,

\(\vec {OA}\) + \(\vec {OC}\) = \(\vec {AC}\) = 2\(\vec {OG}\).............................(1)

[\(\because\) G is a mid-point of AC]

In \(\triangle\)OBD,

\(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\)..........................................(2)

Adding equation (1) and (2)

\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\) + 2\(\vec {OG}\)

∴\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 4\(\vec {OG}\) Hence, Proved

Position vector of P, \(\vec {OP}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\))

Position vector of R, \(\vec {OR}\) = \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))

Position vector of mid-point of \(\vec {PR}\)

= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\)) + \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))]

= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))]

= \(\frac 14\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))

Position vector of Q, \(\vec {OQ}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\))

Position vector of S, \(\vec {OS}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))

Position vector of mid-point of \(\vec {QS}\)

= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\)) + \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))]

= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))]

= \(\frac 14\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))

∴ Position vector of mid-point of \(\vec {PR}\) and \(\vec {QS}\) is same. So, \(\vec {PR}\) and \(\vec {QS}\) bisect each other. Proved

sd

Let, \(\triangle\)ABC be the right angled triangle whose right angle is at A. Let, P be the middle point of the hypotenuse BC.

Take A is the origin and let;

\(\vec {AC}\) = \(\vec c\), \(\vec {AB}\) = \(\vec b\)

\(\vec {AP}\) = \(\vec d\) then \(\vec {AB}\) = \(\vec {AP}\) + \(\vec {PB}\)

[Using triangle law of vector addition]

\(\vec b\) = \(\vec d\) + \(\vec {PB}\) and \(\vec {AC}\) = \(\vec {AP}\) + \(\vec {PC}\)

or, \(\vec c\) = \(\vec d\) - \(\vec {CP}\) i.e. \(\vec c\) = \(\vec d\) - \(\vec {PB}\) [\(\because\) PB = PC]

Now,

It is given that: AB⊥ AC

∴ \(\vec b\) . \(\vec c\) = 0

or, (\(\vec d\) + \(\vec {PB}\)) (\(\vec d\) - \(\vec {PB}\)) = 0

or, d2 - PB2 = 0

or, d2 = PB2

∴ d = PB = PC = AP

∴ The middle point of the hypotenuse of a right-angled triangle is equidistance from the vertices. Proved

Let: ABCD be quadrilateral.

Take A as origin and suppose the position vectors of the vertices B, C, D are \(\vec a\), \(\vec b\) and \(\vec c\) respectively.

Let, P, Q, R and S be the mid-points od AB, BC, CD and DA. PQ, QR, RS and SP are joined.

By using the mid-point formula. We get the position vectors of P, Q, R and S as follows:

\(\vec {AP}\) = \(\frac {\vec o + \vec a}2\) = \(\frac 12\)\(\vec a\)

\(\vec {AQ}\) = \(\frac {\vec a + \vec b}2\) = \(\frac 12\)\(\vec a\) + \(\frac 12\)\(\vec b\)

\(\vec {AR}\) = \(\frac {\vec b + \vec c}2\) = \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)

\(\vec {AS}\) = \(\frac {\vec o + \vec c}2\) = \(\frac 12\)\(\vec c\)

\(\vec {QR}\)

= (position vector of R) - (position vector of Q)

= (\(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)) - (\(\frac 12\)\(\vec a\) + \(\frac 12\) \(\vec b\))

= \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\) - \(\frac 12\)\(\vec b\)

= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(1)

Also,

\(\vec {PS}\)

= (position vector of S) - (position vector of P)

= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(2)

From relation (1) and (2)

\(\vec {QR}\) = \(\vec {PS}\)

∴ QR = PS and QR//PS

∴ PQRS is a parallelogram. Proved

D, E and F are the mid-points of BC, CA and AB.

By using triangle law of vector addition,

\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)

∴ \(\vec {BC}\) = \(\vec {AC}\) - \(\vec {AB}\)

Again,

\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {BD}\)..................................(1)

\(\vec {AD}\) = \(\vec {Ac}\) + \(\vec {CD}\)...................................(2)

Adding (1) and (2)

2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) + \(\vec {BD}\) + \(\vec {CD}\)

or, 2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) +0 [\(\because\) \(\vec {BD}\) = -\(\vec {CD}\)]

∴ \(\vec {AD}\) = \(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\))

\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = -\(\vec {AB}\) + \(\frac 12\)\(\vec {AC}\) = \(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\)

\(\vec {CF}\) = \(\vec {CA}\) + \(\vec {AF}\) = -\(\vec {AC}\) + \(\frac 12\)\(\vec {AB}\) = \(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)

Now,,

\(\vec {AD}\) +\(\vec {BE}\) +\(\vec {CF}\)

=\(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\)) +\(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\) +\(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)

= \(\vec {AB}\) + \(\vec {AC}\) - \(\vec {Ab}\) - \(\vec {AC}\)

= 0

∴\(\vec {AD}\) + \(\vec {BE}\) + \(\vec {CF}\) = 0 Hence, Proved

Let: the position vector of A, B and C are \(\vec a\), \(\vec b\) and \(\vec c\) respectively. \(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {OC}\) = \(\vec c\).

M is the mid-point of AB.

Position vector of M (\(\vec {OM}\)) = \(\frac {\vec {OA} + \vec {OB}}2\) = \(\frac {\vec a + \vec b}2\)

Position vector of N(\(\vec {ON}\)) = \(\frac {\vec {OB} + \vec {OC}}2\) = \(\frac {\vec b + \vec c}2\)

Position vector of MN (\(\vec {MN}\)) = \(\vec {ON}\) - \(\vec {OM}\)

or, \(\vec {MN}\) = \(\frac {\vec b + \vec c}2\) - \(\frac {\vec a + \vec b}2\)

or, \(\vec {MN}\) = \(\frac {\vec b + \vec c - \vec a - \vec b}2\)

or, \(\vec {MN}\) = \(\frac {\vec c - \vec a}2\)

∴ 2\(\vec {MN}\) = \(\vec c\) - \(\vec a\)........................................(1)

\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) = \(\vec c\) - \(\vec a\).........................(2)

From (1) and (2)

\(\vec {AC}\) = 2\(\vec {MN}\) Hence, Proved

Let: O be the centre of a circle and AB be the diameter of the circle.

Let: C be any point in the semi-circle.

Let: \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OC}\) be the position vectors of A, B and C respectively with respect to the origin O.

Using triangle law of vector addition,

\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) and

\(\vec {BC}\) = \(\vec {OC}\) - \(\vec {OB}\)

∴ \(\vec {AC}\) . \(\vec {BC}\)

= (\(\vec {OC}\) - \(\vec {OA}\)) (\(\vec {OC}\) - \(\vec {OB}\))

= \(\vec {OC}\) . \(\vec {OC}\) - \(\vec {OC}\) . \(\vec {OB}\) - \(\vec {OA}\) . \(\vec {OC}\)+ \(\vec {OA}\) . \(\vec {OB}\) [\(\because\) \(\vec {OA}\) = - \(\vec {OB}\)]

= \(\vec {OC}^2\) + \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OB}^2\)

= \(\vec {OC}^2\) - \(\vec {OB}^2\)

= 0 [\(\because\) \(\begin {vmatrix} OC\\ \end {vmatrix}\) = \(\begin {vmatrix} OB\\ \end {vmatrix}\)]

∴ \(\vec {AC}\)⊥ \(\vec {BC}\)

Hence, \(\angle\)ACB = 90° Proved

dasf

Let: ABCD be a parallelogram and M be the mid-point of the diagonal DB. Let: \(\vec {OA}\), \(\vec {OB}\), \(\vec {OC}\) and \(\vec {OD}\) be the position vectors of the points A, B, C and D of the parallelogram with reference to origin O.

Using mid-point theorem;

We have,

\(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OD}\))

or, \(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OA}\) + \(\vec {AD}\)) [\(\because\) \(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\)]

or, \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {BC}\)) [\(\because\) \(\vec {AD}\) = \(\vec {BC}\)]

∴ \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OC}\) which is the position vector of the mid-point of \(\vec {AC}\)

∴ M is the mid-point of AC and BC.

∴ AC and BD bisect each other. Proved

s

Let: \(\vec {OA}\) = \(\vec a\) , \(\vec {OB}\) = \(\vec b\), \(\vec {OC}\) = \(\vec c\) anf \(\vec {OG}\) = \(\vec g\)

The medians AE, BF and CD intersect at G. G is a centroid of the \(\triangle\)ABC.

Using mid-point formula,

In \(\triangle\)BOC,

\(\vec {OE}\) = \(\frac {\vec {OB} + \vec {OC}}2\).......................(1)

In any triangle, the centroid divides median in the ratio of 2 : 1.

In \(\triangle\)AOE,

\(\vec {OG}\) = \(\frac {1 . \vec {OA} + 2 . \vec {OE}}{1 + 2}\) [\(\because\) m : n = 2 : 1]

or, \(\vec g\) = \(\frac {\vec a + 2 . (\frac {\vec {OB} + \vec {OC}}2)}3\) [\(\because\) From eqn (1)]

or, \(\vec g\) = \(\frac {\vec a + \vec b + \vec c}3\)

∴\(\vec g\) = \(\frac13\)(\(\vec a\) + \(\vec b\) + \(\vec c\)) Proved

sdf

Let: PQR is an isosceles triangle where PQ = PR. M is the mid-ppointof QR.

Vertex P and M are joined.

To prove: PM⊥ QR

Using triangle law of vector addition:

\(\vec {QR}\) = \(\vec {QP}\) + \(\vec {PR}\)

\(\vec {QR}\) = - \(\vec {PQ}\) + \(\vec {PR}\) [\(\because\) \(\vec {QP}\) = - \(\vec {PQ}\)]

\(\vec {PM}\) = \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR})\) [\(\because\) M is the mid-point of QR]

Now,

\(\vec {PM}\) . \(\vec {QR}\)

= \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR}\)) (-\(\vec {PQ}\) + \(\vec {PR}\))

= \(\frac 12\) (\(\vec {PR}^2\) - \(\vec {PQ}^2\))

= \(\frac 12\) (PR2 - PQ2)

= \(\frac 12\) (PQ2 - PQ2) [\(\because\) PR = PQ]

= \(\frac 12\)× 0

= 0

Hence, PM⊥ QR.Proved

sdf

Let: PQR is an isosceles triangle.

PM⊥ QR and PQ = PR

To prove: QM = MR

Using triangle law of vector addition,

In \(\triangle\)PQM,

\(\vec {PM}\) = \(\vec {PQ}\) + \(\vec {QM}\)........................(1)

In \(\triangle\)PRM,

\(\vec {PM}\) = \(\vec {PR}\) + \(\vec {RM}\)..........................(2)

From relation (1) and (2)

\(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PR}\) + \(\vec {RM}\)

or, \(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PQ}\) + \(\vec {RM}\) [\(\because\) \(\vec {PQ}\) = \(\vec {PR}\)]

∴\(\vec {QM}\) =\(\vec {RM}\)

Hence, PM bisect the line QR. Proved

wefr

Let: ABC is an isosceles triangle. CD and BE are medians. AB = AC

To prove: BE = CD

Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec p\) and \(\vec {CE}\) = \(\vec {EA}\) = \(\vec q\)

Using triangle law of vector addition,

In \(\triangle\)BCE,

\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\)

\(\vec {BE}\) = \(\vec p\) + \(\vec p\) - \(\vec {EA}\) = 2\(\vec p\) - \(\vec q\).......................(1)

Squaring on both sides,

\(\vec {BE}^2\) = (2\(\vec p\) - \(\vec q\))2

\(\vec {BE}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec q^2\)

\(\vec {BE}^2\) =4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]

\(\vec {BE}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)

In \(\triangle\)BCD,

\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec q\) + \(\vec q\) - \(\vec {DA}\) = 2\(\vec q\) - \(\vec p\)...................................(2)

Squaring on both sides,

\(\vec {CD}^2\) = (2\(\vec q\) - \(\vec p\))2

\(\vec {CD}^2\) = 4\(\vec q^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\)

\(\vec {CD}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]

\(\vec {CD}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)

From above relation,

\(\vec {BE}^2\) = \(\vec {CD}^2\)

or, \(\vec {BE}\) = \(\vec {CD}\)

∴ BE = CD Proved

sdf

Given: BE and CD are median of the \(\triangle\)ABC.

Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec a\) , \(\vec {AE}\) = \(\vec {EC}\) = \(\vec b\)

To prove: AB = AC

In \(\triangle\)BAE,

\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = 2\(\vec a\) + \(\vec b\)

In \(\triangle\)ACD,

\(\vec {DC}\) = \(\vec {DA}\) + \(\vec {AC}\) = \(\vec a\) + 2\(\vec b\)

From question,

\(\vec {BE}\) = \(\vec {DC}\)

or, \(\vec {BE}^2\) = \(\vec {DC}^2\)

or, (2\(\vec a\) + \(\vec b\))2 = (\(\vec a\) + 2\(\vec b\))2

or, 4\(\vec a^2\) + 2\(\vec a\)\(\vec b\) + \(\vec b^2\) = \(\vec a^2\) + 4\(\vec a\)\(\vec b^2\) + 4\(\vec b^2\)

or, 4\(\vec a^2\) - \(\vec a^2\) + 4\(\vec a\)\(\vec b\) - 4\(\vec a\)\(\vec b\) = 4\(\vec b^2\) - \(\vec b^2\)

or, 3\(\vec a^2\) = 3\(\vec b^2\)

or, \(\begin {vmatrix} \vec a^2\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b^2\\ \end {vmatrix}\)

or,\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)

or, 2\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = 2\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)

∴AB = AC

Hence, ABC is an isosceles triangle. Proved

dfc

Let: O be the centre of the circle.

Join OA and OB.

Using triangle law of vector addition,

In \(\triangle\)AOC,

\(\vec {OA}\) = \(\vec {OC}\) + \(\vec {CA}\)

Squaring on both sides,

\(\vec {OA}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))2 ................................(1)

In \(\triangle\)BOC,

\(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)

\(\vec {OB}\) = \(\vec {OC}\) - \(\vec {CA}\) [\(\because\) \(\vec {CB}\) = -\(\vec {CA}\)]

Squaring on both sides,

\(\vec {OB}^2\) = (\(\vec {OC}\) - \(\vec {CA}\))2 ...............................(2)

Subtracting eqn (2) from (1)

\(\vec {OA}^2\) - \(\vec {OB}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))2-(\(\vec {OC}\) - \(\vec {CA}\))2

or, \(\vec {OA}^2\) - \(\vec {OA}^2\) = \(\vec {OC}^2\) + \(\vec {CA}^2\) + 2\(\vec {OC}\) \(\vec {CA}\) - \(\vec {OC}^2\) - \(\vec {CA}^2\) + 2\(\vec {OC}\)\(\vec {CA}\)

or, 0 = 4\(\vec {OC}\)\(\vec {CA}\)

or, \(\vec {OC}\)\(\vec {CA}\) = 0

∴ \(\vec {OC}\)⊥ \(\vec {CA}\)

Hence, OC is perpendicular to AB. Proved

Let, PQR is a right angled triangle.

To prove:PR2 = PQ2 + QR2

Using triangle law of vector addition:

\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)

Squaring on both sides;

\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))2

or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)

or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 × 0 + \(\vec {QR}^2\) [\(\because\) \(\vec {PQ}\) . \(\vec {QR}\) = 0]

∴ \(\vec {PR}^2\) = \(\vec {PQ}^2\) +\(\vec {QR}^2\) Proved

sad

Let, AB be a straight line cuts x-axis at A(a, 0) and y-axis at B(0. b).

Let: P(x, y) be any point on the line AB.

\(\vec {OA}\) = \(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\)

\(\vec {OB}\) = \(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\)

\(\vec {OP}\) = \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\)

\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) =\(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)

\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\)

\(\vec {AB}\) and \(\vec {AP}\) are parallel so,

\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) = k\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)

\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} -ka\\ kb\\ \end {pmatrix}\)

Taking corresponding elements of the equal vectors,

x - a = -ka

\(\frac xa\) = 1 - k .....................(1)

y = bk

\(\frac yb\) = k ............................(2)

Adding relation (1) and (2)

\(\frac xa\) +\(\frac yb\) = 1 - k + k

∴\(\frac xa\) +\(\frac yb\) = 1 Hence, Proved

Proof:

Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P relative to the origin O respectively.

From Question,

\(\frac {AP}{PB}\) = \(\frac mn\)

AP = \(\frac mn\)PB..........................(1)

Using vector of the above relation,

\(\vec {AP}\) = \(\frac mn\)\(\vec {PB}\)

\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and

\(\vec {PB}\) = \(\vec {OB}\) - \(\vec {OP}\)

Putting the value of \(\vec {AP}\) and \(\vec {PB}\) in relation (1)

\(\vec {OP}\) - \(\vec {OA}\) = \(\frac mn\)(\(\vec {OB}\) - \(\vec {OP})\)

or, n \(\vec {OP}\) - n \(\vec {OA}\) = m \(\vec {OB}\) - m \(\vec {OP}\)

or,m \(\vec {OP}\) +n \(\vec {OP}\) = m \(\vec {OB}\) +n \(\vec {OA}\)

or, \(\vec {OP}\) (m + n) =m \(\vec {OB}\) +n \(\vec {OA}\)

∴ \(\vec {OP}\) = \(\frac {m.\vec {OB} + n.\vec {OA}}{m+n}\) Hence, Proved

Proof:

Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P respectively in relation to the origin O.

From Question,

\(\frac {AP}{PB}\) = \(\frac mn\)

n. AP = m. PB

Using vector of the above relation,

n \(\vec {AP}\) = m \(\vec {PB}\)...........................(1)

From triangle law of vector addition:

\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and

\(\vec {BP}\) = \(\vec {OP}\) - \(\vec {OB}\)

Putting the value of \(\vec {AP}\) and \(\vec {BP}\) in relation (1)

n (\(\vec {OP}\) - \(\vec {OA}\)) = m (\(\vec {OP}\) - \(\vec {OB}\))

or, n . \(\vec {OP}\) - n . \(\vec {OA}\) = m . \(\vec {OP}\) - m . \(\vec {OB}\)

or,n . \(\vec {OP}\) -m . \(\vec {OP}\) =n . \(\vec {OA}\)- m . \(\vec {OB}\)

or,\(\vec {OP}\) (n - m) =n . \(\vec {OA}\)- m . \(\vec {OB}\)

or,\(\vec {OP}\) = \(\frac {n . \vec {OA} - m . \vec {OB}}{n - m}\)

∴\(\vec {OP}\) = \(\frac {m. \vec {OB} - n. \vec {OA}}{m - n}\) Hence, Proved

Given:

OABC is a parallelogram.

\(\vec {CP}\) : \(\vec {OP}\) = \(\vec {CQ}\) : \(\vec {QB}\) = 1 : 3

\(\vec {OA}\) = \(\vec a\) and \(\vec {OC}\) = \(\vec c\)

To prove: \(\vec {PQ}\)//\(\vec {OB}\) and find \(\vec {PQ}\) = ?

From Question:

\(\frac {\vec {CP}}{\vec {PO}}\) = \(\frac 13\) = \(\frac {\vec {CQ}}{\vec {QB}}\)

∴ \(\vec {PO}\) = 3 \(\vec {CP}\) and \(\vec {QB}\) = 3 \(\vec {CQ}\)

Using triangle law of vector addition:

\(\vec {OB}\) = \(\vec {OA}\) + \(\vec {AB}\)

or, \(\vec {OB}\)= \(\vec a\) + \(\vec c\)

or, \(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)

or, \(\vec {OB}\) = \(\vec {OP}\) + \(\vec {PC}\) + \(\vec {CQ}\) + \(\vec {QB}\)

or, \(\vec {OB}\) = - \(\vec {PO}\) + \(\vec {PC}\) + \(\vec {CQ}\) + 3\(\vec {CQ}\)

or, \(\vec {OB}\) = -3 \(\vec {CP}\) - \(\vec {CP}\) + 4 \(\vec {CQ}\) [\(\because\) \(\vec {PO}\) = 3 \(\vec {CP}\)]

or, \(\vec {OB}\) = - 4 \(\vec {CP}\) + 4 \(\vec {CQ}\)

or, \(\vec {OB}\) = 4 \(\vec {PC}\) + 4 \(\vec {CQ}\)

or,\(\vec {OB}\) = 4 (\(\vec {PC}\) + \(\vec {CQ}\))

∴\(\vec {OB}\) = 4 \(\vec {PQ}\)

∴\(\vec {OB}\)//\(\vec {PQ}\) [\(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]

and \(\vec {PQ}\) = \(\frac 14\)\(\vec {OB}\) = \(\frac 14\)(\(\vec a\) + \(\vec c\)) Proved

Given:

ABCD is a parallelogram. L and M are mid-points of sides BC and CD respectively.

Using triangle law of vector addition:

\(\vec {AM}\) = \(\vec {AD}\) + \(\vec {DM}\) = \(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\)...............(1)

\(\vec {AL}\) = \(\vec {AB}\) + \(\vec {BL}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\).....................(2)

Adding (1) and (2)

\(\vec {AM}\) +\(\vec {AL}\) =\(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\) +\(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)

or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {AD} + \vec {DC} + 2 \vec {AB} + \vec {BC}}2\)

or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {BC} + \vec {AB} + 2 \vec {AB} + \vec {BC}}2\) [\(\because\) \(\vec {AB}\) = \(\vec {DC}\) and \(\vec {AD}\) = \(\vec {BC}\)]

or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {3 \vec {AB} + 3 \vec {BC}}2\)

or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\) (\(\vec {AB}\) + \(\vec {BC}\))

∴ \(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\)\(\vec {AC}\)

Hence, \(\vec {AL}\) +\(\vec {AM}\) = \(\frac 32\)\(\vec {AC}\) Proved

0%
  • Find (overrightarrow a).(overrightarrow b), when

    (overrightarrow a) = 2(overrightarrow i) + 3(overrightarrow j) and (overrightarrow b) = -2(overrightarrow i) - (overrightarrow j)

    15


    13


    14


    12


  • Find (overrightarrow a). (overrightarrow b), when

    (overrightarrow a) = 2(overrightarrow i) + 3(overrightarrow j) and (overrightarrow b) = 2(overrightarrow i) - (overrightarrow j)

    2


    -1


    3


    1


  • Find (overrightarrow a). (overrightarrow b), when

    (overrightarrow a) = (overrightarrow i) + 2(overrightarrow j) and (overrightarrow b) = 3(overrightarrow j) - 2(overrightarrow i)

    1


    -2


    2


    -1


  • If (overrightarrow a) and (overrightarrow b) are perpendicular to each other, find the value of m where:

    (overrightarrow a) = 3(overrightarrow i) + m(overrightarrow j) and (overrightarrow b) = 3(overrightarrow i) - 3(overrightarrow j)

    5


    3


    4


    2


  • If (overrightarrow a) and (overrightarrow b) are perpendicular to each other, find the value of m:

    (overrightarrow a) = -4(overrightarrow i) + 7(overrightarrow j) and  (overrightarrow b) = 14(overrightarrow i) - 3m(overrightarrow j)

    2


    6


    (-frac{8}{3})


    8


  • If (overrightarrow a) and (overrightarrow b) are two vectors such that |(overrightarrow a)| = 16 and |(overrightarrow b)| = 3 and  (overrightarrow a).(overrightarrow b) = 24 (sqrt 3), find the angles between (overrightarrow a) and (overrightarrow b).

    400


    600


    300


    450


  • If |(overrightarrow a)| = 4, |(overrightarrow b)| = 6 and θ = 600, find the value of (overrightarrow a). (overrightarrow b).

    11


    12


    13


    10


  • The position vectors of points A and B of a line are (egin{pmatrix}1\3\ end{pmatrix}) and (egin{pmatrix}3\5\ end{pmatrix}) respectively.Find the position vector of mid point M of AB.

    (egin{pmatrix}5\4\ end{pmatrix})


    (egin{pmatrix}2\4\ end{pmatrix})


    (egin{pmatrix}1\4\ end{pmatrix})


    (egin{pmatrix}2\5\ end{pmatrix})


  • If the position vectors of the points A and B are 3(overrightarrow i) + 4(overrightarrow j) and 5(overrightarrow i) - 2(overrightarrow j) respectively. Find the position vector of the mid-point M of AB.

     

    3(overrightarrow i) + (overrightarrow j)


    5(overrightarrow i) + (overrightarrow j)


    4(overrightarrow i) + (overrightarrow j)


    2(overrightarrow i) + 3(overrightarrow j)


  • The position vector of P and Q are 2(overrightarrow i) + 7(overrightarrow j) and 4(overrightarrow i) - 3(overrightarrow j). Find the position vector of a point which divides PQ externally in the ratio of 2:3.

    25(overrightarrow j) - (overrightarrow i)


    27(overrightarrow j) - 2(overrightarrow i)


    29(overrightarrow j) - (overrightarrow i)


    23(overrightarrow j) - 2(overrightarrow i)


  • If the points X(-1, -1), Y(5, 1) and Z(2, 6) are the vertices of triangle XYZ, find the position vector of its centriod.

    (egin{pmatrix}2\3\ end{pmatrix})


    (egin{pmatrix}1\2\ end{pmatrix})


    (egin{pmatrix}2\1\ end{pmatrix})


    (egin{pmatrix}2\2\ end{pmatrix})


  • Find the position vector of a point in the x-axis which divides the line joining the points (2, -1) and (8, 2) in the ratio 1:2.

    (egin{pmatrix}4\2\ end{pmatrix})


    (egin{pmatrix}4\0\ end{pmatrix})


    (egin{pmatrix}2\4\ end{pmatrix})


    (egin{pmatrix}2\2\ end{pmatrix})


  • If (overrightarrow a) + (overrightarrow b) + (overrightarrow c) = 0, |(overrightarrow a)| = 4, |(overrightarrow b)| = 10 and (overrightarrow a).(overrightarrow b) = 30, find the value of |(overrightarrow c)|

    13


    15


    12


    14


  • If P(4, 4), Q(2, 2) and R(4, 2), then find (overrightarrow PQ). (overrightarrow QR).

    -4


    -5


    -3


    -2


  • if A(4, 5), b(5, 2) and C(-2, 3) are the vertices of a triangle ABC then find (overrightarrow AB). (overrightarrow AC).

    two


    five


    zero


    eight


  • You scored /15


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The vector sum of medians of a triangle is always zero

Dhananjay Prasad Shah

Prove by vector method that perpendicular drown from vertices to opposite side of triangle are concurrent

bishnu

sabai note lai apps banayara market ma lyaunus..... suvkamana xa.

Raju

Pqrs is a parallelogram. M And N are the points on diagonal so. Prove that pmrn is a parallelogram.